A 250-kg motorcycle goes around an unbanked turn of radius 13.7 m at a steady 96.5 km/h.What is the magnitude of the sum of the forces exerted on the motorcycle? A)719 N B)2.95 × 10³ N C)1.31 × 10⁴ N D)4.31 × 10⁴ N

Question

Quizplus · Accepted Answer

First, we need to find the centripetal acceleration acting on the motorcycle.

v = 96.5 km/h = 26.8 m/s (convert to m/s) 
r = 13.7 m

a = v^2/r = (26.8 m/s)^2 / 13.7 m = 52.4 m/s^2

Next, we can use Newton's second law to find the force (F) exerted on the motorcycle.

F = ma = (250 kg) x (52.4 m/s^2) = 13,100 N

Lastly, we need to consider any other forces acting on the motorcycle. We can assume that the normal force and the force due to friction are the only other forces present. Since the motorcycle is not banking, the normal force and force due to friction must add up to be equal in magnitude to the centripetal force.

Therefore, the magnitude of the sum of forces is 2F = 2(13,100 N) = 26,200 N or 1.31 × 10^14 N (in scientific notation).

The correct choice is C.

A 250-Kg Motorcycle Goes Around an Unbanked Turn of Radius